IGO 2021 A1

Acute-angled triangle $ABC$ with circumcircle $\omega$ is given. Let $D$ be the midpoint of $AC$ , $E$ be the foot of altitude from $A$ to $BC$ , and $F$ be the intersection point of $AB$ and $DE$ . Point $H$ lies on the arc $BC$ of $\omega$ (the one that does not contain $A$ ) such that $\angle BHE=\angle ABC$ . Prove that $\angle BHF=90^\circ$ .